Diagonalizing the ‘restricted’ Hilbert transform on $L^2(0,1)$, $f(z_1) \mapsto \mathrm{p.v.} \int_0^1 \frac{i}{z_1-z_2}f(z_2) dz_2$ Consider the following operator on functions $\mathcal{T}: L^2(0,1) \to  L^2(0,1)$ over the complex numbers.
\begin{equation}
(\mathcal{T} f)(z_1) = \mathrm{p.v.} \int_0^1 \frac{i}{z_1-z_2}f(z_2) dz_2
\end{equation}
where $\mathrm{p.v.}$ means the Cauchy principal value. This is an analogue of the Hilbert Transform except restricted to functions on an interval.
Note that this operator is Hermitian, so there should be an orthogonal basis of eigenfunctions. Is there a known description of such an orthogonal basis of the eigenfunctions and eigenvalues of this operator and a relevant 'Fourier inversion' formula?
 A: This was carried out in
Koppelman, W.; Pincus, J. D., Spectral representations for finite Hilbert transformations, Math. Z. 71, 399-407 (1959). ZBL0085.31701.
The spectrum is purely continuous on $[-\pi,\pi]$ (with the OP's normalisation) and Theorem 3.1 gives the explicit Fourier-Plancherel theorem that diagonalises this operator (see Theorem 3.2). Note that the Hilbert transform should not be expected to be compact (it is not an integral operator due to the presence of the principal value) and so the fact that there are no eigenfunctions should not be surprising (this can presumably also be shown by some complex analysis argument, though I haven't worked out the details).  It is also observed that the same Fourier-type transform diagonalises the self-adjoint differential operator $i \sqrt{x(1-x)} \frac{d}{dx} \sqrt{x(1-x)}$ in addition to the finite Hilbert transform (in particular, this differential operator commutes with that transform, and in fact can be expressed using the functional calculus of that transform).
